Question

Find three rational numbers between 4 and 4 whole 2 by 3

Answer 1

$4 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 4 \frac{2}{3} \\ \\ 4 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \frac{14}{3}$

Here,

$x = 4, \: y = \frac{14}{3} , \: n = 3$

Now,

$d = \frac{y - x}{n + 1} \\ \\ d = \frac{( \frac{14}{3} - 4 )}{3 + 1} = \frac{( \frac{14 - 12}{3} )}{4} = \frac{( \frac{2}{3} )}{4} = \frac{2}{12} = \frac{1}{6}$

Hence,

$1st \: \: Number = (x + d) = (4 + \frac{1}{6} ) = \frac{25}{6} \\ \\ \\ 2nd \: \: Number = (x + 2d) = (4 + \frac{2}{6}) =( 4 + \frac{1}{3} ) = \frac{13}{3} \\ \\ \\ 3rd \: \: Number = (x + 3d) = (4 + \frac{3}{6}) = (4 + \frac{1}{2}) = \frac{9}{2}$